If a, m and n are positive integers, is n²ª a multiple of mª?(1) n is a multiple of m/2(2) n is a multiple of 2m

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.D. Either statement BY ITSELF is sufficient to answer the question.E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

(B) Statement (1) tells us that (m/2) is a divisor of n. Suppose m = 6. Then n can be 3. In this case n²ª is 3²ª = 9ª and is clearly NOT a multiplier of 6ª.Or n can be 6. In this case n²ª is 6²ª and is clearly a multiplier of 6ª. Therefore the statement (1) is NOT sufficient by itself to answer the question.

Statement (2) tells us that (2m) is a divisor of n. Therefore m is a divisor of n as well. So n²ª is a multiple of m²ª, which entails n²ª to be a multiple of mª. The statement (2) is sufficient by itself to answer the question.

You should always specify what exactly you don't understand about a question. Is it some specific logical reasoning? Or is it some method used to solve a question? So if you don't find the answer to you question in this topic, please, specify it.

Let us briefly go through analysis of this question once again.

The question asks "Is n²ª a multiple of mª?", which in other words means "Is is n²ª = k × mª?", where k is some positive integer.

The second statement tells us that n is a multiple of 2m, which in other words means n = r × 2m, where r is some positive integer. Let us raise the equality to 2a-th power.

Now, let's use the first statement by itself.The first statement tells us that n is a multiple of m/2, which in other words means n = r × (m/2) = , where r is some positive integer and m/2 is an integer as well. Let us raise the equality to 2a-th power.

n²ª = (r × m/2)²ª = r²ª × m²ª / 2²ª = (r²ª × mª / 2²ª) × mª.

If r²ª × mª / 2²ª is an integer, then n²ª = k × mª and so n²ª is a multiple of mª.If r²ª × mª / 2²ª is NOT an integer, then n²ª does NOT equal k × mª (where k is an integer) and son²ª is NOT a multiple of mª.

(r²ª × mª / 2²ª) is an integer if (r² × m) is divisible by 4, which is clearly can be TRUE as well as can be NOT TRUE. (plugged in values in the explanation prove it).

Therefore the second statement by itself is sufficient, while the first one is NOT.

I got this wrong, and thought about it for some time. Here is my explanation:Given that a, m and n are positive integers, we need to find out if n²ª is of the form kmª.i.e. is n²ª/mª = k where k is any positive integer.This reduces to (n²/m)ª = k or in other words it is sufficient to determine if n²/m is an integer.The first clue given is that n = am/2, where a is an integerin that case n² = (a²)m²/4 and n²/m = ((a²)m²)/(4m) = ((a²)m)/4 which is an integer, only if (a²)m is a multiple of 4.Since no other conditions are given to bind a,or m, this need not be necessary for all combinations of a and m hence insufficient (a = 1, m = 3 not true, a = 2, m = 1, true)The second clue has already been proved as sufficient so the answer is "B"!!

In fact, the answer is B. So your reasoning leads to the correct answer.

But I'd like to bring your attention to the following lines in your reasoning:

1. "This reduces to (n²/m)ª = k or in other words it is sufficient to determine if n²/m is an integer."

When you made this conclusion, you had to realize that n²/m is a rational number by definition (= p/t), it is either an irreducible fraction or an integer. It can not be a fraction, because then (n²/m)ª = (p/t)ª = pª/tª, which is NOT an integer and therefore can NOT equal k. So, as you conclude, n²/m is an integer.

Note, that IF n²/m could be an irrational number, e.g. ²√k, then this reasoning would NOT be true.

2. "The first clue given is that n = am/2, where a is an integer"Here, and in the lines that follow, the variable a is NOT a good notation, since it can be easily confused with the variable a in the question statement.

3. "(a = 1, m = 3 not true, a = 2, m = 1, true)"Statement 1 implements that m is even, since m/2 must be an integer.Why must m/2 be an integer? Because the term "multiple" is defined for integers only.

Therefore m can NOT equal 3 or 1.Still, n²/m = ((f²)m)/4 is an integer, when f = 1, m = 4; and is NOT an integer, when f = 1, m = 2.

Overall, your reasoning led to a correct answer, but there are some inaccuracy moments it it. Those moments could have led to a wrong choice, if the question was a different one, so be very careful in the future.

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